Answer
The solutions are $x=1\pm\dfrac{\sqrt{6}}{2}$
Work Step by Step
$2x^{2}-4x-1=0$
Take out common factor $2$ from the left side:
$2\Big(x^{2}-2x-\dfrac{1}{2}\Big)=0$
Take the $2$ to divide the right side:
$x^{2}-2x-\dfrac{1}{2}=\dfrac{0}{2}$
$x^{2}-2x-\dfrac{1}{2}=0$
Take $\dfrac{1}{2}$ to the right side:
$x^{2}-2x=\dfrac{1}{2}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-2$:
$x^{2}-2x+\Big(\dfrac{-2}{2}\Big)^{2}=\dfrac{1}{2}+\Big(\dfrac{-2}{2}\Big)^{2}$
$x^{2}-2x+1=\dfrac{1}{2}+1$
$x^{2}-2x+1=\dfrac{3}{2}$
Factor the left side of the equation, which is a perfect square trinomial:
$(x-1)^{2}=\dfrac{3}{2}$
Take the square root of both sides:
$\sqrt{(x-1)^{2}}=\pm\sqrt{\dfrac{3}{2}}$
$x-1=\pm\dfrac{\sqrt{6}}{2}$
Solve for $x$:
$x=1\pm\dfrac{\sqrt{6}}{2}$
